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February 27th, 2014, 03:02 PM  #1 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  The mystery of cos 15/sr280
This is my first post on a mathematical forum, and I hope readers can work out my puzzle. Lay off twice the diameter of a circle on a slope equal to cos 15/(square root of 280) and the vertical rise is the side of a square approximately equal to the area of the circle: it is equal to the area calculated using the pi approximation 22/7. The cosine of 15/sr280 is 26 degrees 18 minutes 30 seconds to the nearest arc minute. Inside the Great Pyramid of Giza is a puzzling structure known as the Grand Gallery, which is a narrow passageway with soaring elevations. In the nineteenth century its sloping floor was found to be 26 degrees 18 minutes within 2 arc minutes. The Grand Gallery was built perpendicular to its slope, and the intended perpendicular height near the entrance appears to have been exactly 15 royal cubits. Therefore the intended vertical height may have been the square root of 280 in royal cubits, which is approximately 16 royal cubits 20 1/2 digits, with 28 digits per royal cubit. (If two parallel lines make an angle of less than 90 degrees to the horizontal then a rightangled triangle can be placed between the parallel lines such that perpendicular separation is base of the triangle and the vertical separation is the hypotenuse of the triangle, with the angle between these sides equal to the slope.) The original design height of the Great Pyramid was 280 royal cubits, as determined from its base sidelength of 440 royal cubits and the slope of the faces. A circle with a diameter of 280 royal cubits is approximately equal to the area of the 'pyramid triangle' , and may have been seen as exactly equal if the pi approximation 22/7 was regarded as exact. Laying off twice the diameter of 280 royal cubits on a slope equal to cos 15/sr280 results in a vertical rise equal to the square root of 61.600 square royal cubits, so the square is 61,600 square royal cubits. The area of the pyramid triangle is also 61,600 square royal cubits (1/2 x 440 x 280). The reader may now work out the mystery of cos 15/sr280. The mystery is 'Why choose a slope where one has to lay off '2 x diameter of a circle' in order to get the sidelength of the square? A simple answer is that it results in a slope which could be used as a passageway, whereas laying off say '1 x diameter of a circle' would be too steep. But there is another reason, presumably spotted by the architect, which is an elegant mathematical representation of 'squaring the circle'. The answer is given in appendix H and appendix J of my website http://www.ancientcalendar.co.uk Mark 
February 27th, 2014, 05:01 PM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: The mystery of cos 15/sr280 Quote:
But wait... you're measuring the cosine in degrees? The cosine's argument should be in degrees (or radians or gradiants), but not its output. What's going on here?  
February 28th, 2014, 10:33 AM  #3 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  Re: The mystery of cos 15/sr280
The ancient Egyptians did not need cosines or degrees. I attempted to express the problem in cosines and degrees so it would be easier to grasp the puzzle. A slope with a rise of sr55, a run of 15, and a hypotenuse of sr280 is all you need to consider. But for my education would it be correct to write: 'the angle calculated from the cosine 15/sr280 is approximately 26 degrees 18 minutes 30 seconds' ? The cosine is 0.89642... on my calculator and 'inverse cos' converts to 26.308.. degrees I used tables when I was at school. My casio fx110 scientific calculator from 1978 still seems like magic to me. You don't need either a calculator or tables to solve the puzzle. I hope this clears up the confusion. Mark 
February 28th, 2014, 11:02 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 931 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: The mystery of cos 15/sr280 Quote:
> cos((26+18.5/60)*Pi/180) $\color{#BF0000}{\small0.896421}$97880903125558142 75478056130615432873793510338723230708941919565 > 15/sqrt(280) $\color{#BF0000}{\small0.896421}$45700079522997661 28847698437386351593242485773569405623675861630 Last edited by skipjack; October 14th, 2016 at 08:05 PM.  
February 28th, 2014, 12:59 PM  #5 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  Re: The mystery of cos 15/sr280
Interestingly, the intended slope length of the floor of the Grand Gallery was probably 88 royal cubits, and the intended vertical rise was probably 39 royal cubits. Sine 39/88 = sine 0.443.. inverse sine 0.443 = 26.307.. degrees, which is approximately 26 degrees 18 minutes 25 seconds The observed slope is likely to be within 3 arc minutes of the intended slope if the build standard was accurate to 1 part in 1000, and assuming the structure has not subsided since it was built in the third millennium BC. Theoretical angles should be compared to measured angles. For example, a slope intended to be a rise of 1 for a run of 2 is likely to be within 3 arc minutes of the approximate theoretical angle, which is 26 degrees 34 minutes to the nearest arc minute. The ancient Egyptians may have made templates of many triangles. Sine 39/88 may have appeared to be the same slope as sine sr55/sr280, as the difference is a small fraction of an arc minute. Mark 

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